1
$\begingroup$

I need to compute second distributional derivative of the function

$$ g(x) = cos|x-2|, $$

but I'm not sure about my solution.

\begin{align} \left<g'', \varphi \right> = \left<g, \varphi''\right> &= \int^{\infty}_{-\infty} \underbrace{cos|x-2|}_u \cdot \underbrace{\varphi''(x)}_{v'} dx \\ &= \Big| \begin{array}{@{}cc@{}} u=cos|2-x| & u'=sin(2-x) \\ v = \varphi'(x) & v'= \varphi''(x) \end{array} \Big|\\ &=\underbrace{\left[ cos|x-2| \cdot \varphi'(x) \right]^\infty_{-\infty}}_0 - \int^\infty_{-\infty} \underbrace{sin(2-x)}_u \cdot \underbrace{\varphi'(x)}_{v'} dx\\ &= \Big| \begin{array}{@{}cc@{}} u=sin(2-x) & u'=- cos(2-x) \\ v = \varphi(x) & v'= \varphi'(x) \end{array} \Big|\\ &= - \underbrace{\left[ sin(2-x) \cdot \varphi(x) \right]^\infty_{-\infty}}_0 + \int^\infty_{-\infty} - cos(2-x) \cdot \varphi(x) dx \end{align}

So, is $- cos(2-x)$ the correct distributional derivative?

$\endgroup$
  • $\begingroup$ @JeanMarie So you're telling me to divide integral into multiple intervals, where cosine function behaves differently? I thought, that cos|x-2| behaves exactly like cos(x), just shifted to the right... desmos.com/calculator/djg73s8yoi $\endgroup$ – Eenoku Apr 23 '16 at 22:33
  • $\begingroup$ I am sorry, I withdraw my remark: I had misread $sin|x-2|$ as $|sin(x-2)|$ ... I will even cancel it, not for hiding my misreading, but for readers not being puzzled... $\endgroup$ – Jean Marie Apr 23 '16 at 22:39
  • $\begingroup$ @JeanMarie It's ok, no problem... Just, what's your opinion now? Is it ok? :-) $\endgroup$ – Eenoku Apr 23 '16 at 22:41
  • $\begingroup$ @JeanMarie Thank you very much! Could you, please, add this as an answer, so I could accept it? $\endgroup$ – Eenoku Apr 23 '16 at 22:56
  • $\begingroup$ All right. Nice of you. I will precise something, that I had overlooked $\endgroup$ – Jean Marie Apr 23 '16 at 23:16
1
$\begingroup$

Your computations are exact but you can bypass them.

In fact, there is a general result that says that, till its $n$th derivative (including it), the distributional derivatives and the ordinary derivatives of a $C^n$ function coincide.

In this case where in fact we deal with a $C^{\infty}$ function ($\cos(|x-2|)$ is identical with $\cos(x-2)$), there will be no problem at any order of derivation.

$\endgroup$
  • 1
    $\begingroup$ As you see, I have modified my explanation: it is necessary to work with locally integrable functions but it is not sufficient: such a function can be discontinuous, generating diracs which definitely would perturbate the equivalence between distributional and ordinary derivatives. $\endgroup$ – Jean Marie Apr 23 '16 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.